Hi. I'm a first year student of electronics engineering. I am stuck on a question. We have a 1600W system operating on 12V DC. How many batteries are required to operate the system for 2 days? Use real battery specifications in your calculations.

Here's what I've come up with so far:

Let's take a 12V battery with the following current specs:
20-hr rate Capacity to 1.75VPC at 20°C 24 Ah (what I understood from this is that the battery will be able to provide 24/20=1.2A per hour for 20 hours).

Since the system is 1600W and operates at 12V it needs 1600/12=133.33A current.

First of all I don't understand how to convert Amp/hour to Amp. I realize that I need to connect the batteries in parallel because I have the required voltage but need to increase the current. But how do I calculate how many batteries I need? Do I randomly keep adding the batteries and calculating until I reach 133.33A?

Next, I need to run the system on batteries for 2 days so what effect will that have on calculations?

I'm sorry if this isn't making much sense. I hope someone can give me a clue about what to do.

Hi. I'm a first year student of electronics engineering. I am stuck on a question. We have a 1600W system operating on 12V DC. How many batteries are required to operate the system for 2 days? Use real battery specifications in your calculations.Here's what I've come up with so far:Let's take a 12V battery with the following current specs:20-hr rate Capacity to 1.75VPC at 20°C 24 Ah (what I understood from this is that the battery will be able to provide 24/20=1.2A per hour for 20 hours).Since the system is 1600W and operates at 12V it needs 1600/12=133.33A current.First of all I don't understand how to convert Amp/hour to Amp. I realize that I need to connect the batteries in parallel because I have the required voltage but need to increase the current. But how do I calculate how many batteries I need? Do I randomly keep adding the batteries and calculating until I reach 133.33A? Next, I need to run the system on batteries for 2 days so what effect will that have on calculations?I'm sorry if this isn't making much sense. I hope someone can give me a clue about what to do.Thanks

First, some basics:It's amp-hour, not amp/hr. Very important to get that straight.Second, a watt is a rate of energy consumption or delivery. You want to compute a total energy.So compute how much energy is required. Then you should be able to figure out the batteries needed.

My only quibble is with "power consumed". If anything, it is energy that is "consumed" (although that is itself problematic if taken too literally). It would be better to say that power is the rate at which energy is consumed (or delivered), so that energy is therefore the time integral of power. If power is constant, then the integral just becomes energy = power x time. If power is in watts, and time is in seconds, energy will be in joules (not Joules; two Joules are two members of the Joule family; "watts" for the same reason -- but we capitalize the abbreviations of units named for people).

So a 1600W system would need 1600*48=76800Wh Joules of energy for 2 days.

A watt-hour isn't a joule, although both are measures of energy; there is no "Wh joule" unit. It would be a good idea for you to sit with a nice cup of tea and carefully review the formal definitions of the various units. That by itself would be enormously helpful to you in sorting out the concepts.

A battery capable of providing 24Ah for 20 hours can provide 20/48=0.416Amp for 48 hours.

To keep straight that energy is a total amount, and not a rate, perhaps it would help to think of "24Ah" as "24 dollars" in your pocket or bank vault. Then you can see that "24Ah for 20 hours" makes as much sense as "I have $24 in my pocket for 20 hours." Your maths comes out wrong in consequence.

The power of the battery will be 12*0.416=4.992W.

Similarly, your power calculation is off because you've got the wrong current.

Again, I recommend strongly that you not attempt any calculations at all until you have sorted out the definitions of the quantities you are working with. It's not difficult, but you have to decide to do it, and then do it. I know that it's seductive to dive into the calculations as a shortcut, but experimentally, we see that it isn't working too well.

To keep straight that energy is a total amount, and not a rate, perhaps it would help to think of "24Ah" as "24 dollars" in your pocket or bank vault. Then you can see that "24Ah for 20 hours" makes as much sense as "I have $24 in my pocket for 20 hours." Your maths comes out wrong in consequence.

I wrote 24Ah for 20 hours because the datasheet provides 2 Ah values. One for 10 hours and another for 20 hours. I don't understand how to calculate current. Could you give me an idea? The datasheet of the battery I'm talking about is linked below.

To keep straight that energy is a total amount, and not a rate, perhaps it would help to think of "24Ah" as "24 dollars" in your pocket or bank vault. Then you can see that "24Ah for 20 hours" makes as much sense as "I have $24 in my pocket for 20 hours."

That is actually a common term in batteries. It means that if you have a discharge that lasts for 20 hours, the battery will deliver 24 amp hours. (If the discharge is faster, say 10 hours, then more voltage is lost to resistance, and total capacity to cutoff voltage might be 22 amp hours.)

You have a 1600W system operating on a 12V battery. So your current draw is (1600/12) amps. The maximum discharge current for that battery is 48 amp hours. So your first question is - can one battery supply the current draw?

If no, keep in mind that as you put batteries in parallel, their current adds. (When you put them in series, their voltage adds.) So how many batteries would you need in parallel to supply your load at all?

Next you want to find out how many batteries you need to operate it for 48 hours. 1 amp for 48 hours would be 48 amp-hours, and those batteries are 24ah. So that means you'd use up the entire battery in 24 hours at a 1 amp draw. What can you do? You can put a battery in parallel; amp-hours add when you parallel them so now you have 48 amp hours of battery (2 x 24ah.) If you need to draw 2 amps, now you need 4 batteries. Etc etc.

As for the 22 vs 24 amp hour thing - since your discharge will be a 48 hour discharge (which is greater than 20 hours) the 20 hour number applies.

If you want some extra credit:

In lead acid battery designs you never parallel more than 3-4 strings; the batteries lose balance and die quickly. So if you end up with something like 20 batteries in parallel you might point this out, and suggest either larger batteries (i.e. 400 ah batteries) or a higher voltage (so you can do it with fewer batteries in parallel.)

It is also a rule that you never plan to discharge more than 50% of a lead acid battery, since discharging it to 100% more than a few times will destroy it. So you might point out that a reliable system of this type will require twice the amp-hours that it seems like on paper.

To keep straight that energy is a total amount, and not a rate, perhaps it would help to think of "24Ah" as "24 dollars" in your pocket or bank vault. Then you can see that "24Ah for 20 hours" makes as much sense as "I have $24 in my pocket for 20 hours."

That is actually a common term in batteries. It means that if you have a discharge that lasts for 20 hours, the battery will deliver 24 amp hours. (If the discharge is faster, say 10 hours, then more voltage is lost to resistance, and total capacity to cutoff voltage might be 22 amp hours.)

Surely you must be aware that introductory textbook problems are abstractions precisely designed to build a student's understanding step by step. Of course capacity is indeed a function of discharge rate, thanks to internal resistance. But that has nothing to do with this student's problem. It is idealized to the point where internal resistance -- and its nonlinearity -- are not part of the problem statement. So I properly flagged his/her use of the term as improper in this context. To bring up your tidbit will merely confuse this student.

Hi. I'm a first year student of electronics engineering. I am stuck on a question. We have a 1600W system operating on 12V DC. How many batteries are required to operate the system for 2 days? Use real battery specifications in your calculations.

Here's what I've come up with so far:

Let's take a 12V battery with the following current specs:
20-hr rate Capacity to 1.75VPC at 20°C 24 Ah (what I understood from this is that the battery will be able to provide 24/20=1.2A per hour for 20 hours).

Since the system is 1600W and operates at 12V it needs 1600/12=133.33A current.

First of all I don't understand how to convert Amp/hour to Amp. I realize that I need to connect the batteries in parallel because I have the required voltage but need to increase the current. But how do I calculate how many batteries I need? Do I randomly keep adding the batteries and calculating until I reach 133.33A?

Next, I need to run the system on batteries for 2 days so what effect will that have on calculations?

I'm sorry if this isn't making much sense. I hope someone can give me a clue about what to do.

The datasheet says nothing of the kind. You're not reading the datasheet as written, so that's another problem to be addressed.

It says that the capacity is 24Ah or 22.23Ah, depending on whether you discharge the battery over a 20 or 10 hour period, respectively. You see that the capacity changes slightly, but ignore that (and billvon's related comment) for now. You've got more fundamental problems to clear up first. So, assume for now that the battery capacity is 24amp-hours.

One for 10 hours and another for 20 hours. I don't understand how to calculate current. Could you give me an idea?

Again, I urge you to sit down and study the definitions of the quantities involved. Here are some questions to guide you:

1) Is an amp-hour a unit of energy? If not, what is the meaning of the unit?

2) Why is an amp-hour used as a measure of capacity? (hint: consider why voltage is seemingly ignored here)

The maximum discharge current for that battery is 48 amp hours. So your first question is - can one battery supply the current draw?

No

Originally Posted by billvon

Next you want to find out how many batteries you need to operate it for 48 hours. 1 amp for 48 hours would be 48 amp-hours, and those batteries are 24ah. So that means you'd use up the entire battery in 24 hours at a 1 amp draw. What can you do? You can put a battery in parallel; amp-hours add when you parallel them so now you have 48 amp hours of battery (2 x 24ah.) If you need to draw 2 amps, now you need 4 batteries. Etc etc.

the system needs (1600/12)*48=6400Ah energy to run for 2 days straight. For a battery capable of giving 24Ah I'll need (6400/24)=267 such batteries in parallel.

That was surprisingly simple

Originally Posted by billvon

As for the 22 vs 24 amp hour thing - since your discharge will be a 48 hour discharge (which is greater than 20 hours) the 20 hour number applies.

Does this mean the battery can supply 24/48=0.5A for 48 hours? or will we be able to get 24/20=1.2A for 48 hours?

Originally Posted by billvon

If you want some extra credit:

In lead acid battery designs you never parallel more than 3-4 strings; the batteries lose balance and die quickly. So if you end up with something like 20 batteries in parallel you might point this out, and suggest either larger batteries (i.e. 400 ah batteries) or a higher voltage (so you can do it with fewer batteries in parallel.)

It is also a rule that you never plan to discharge more than 50% of a lead acid battery, since discharging it to 100% more than a few times will destroy it. So you might point out that a reliable system of this type will require twice the amp-hours that it seems like on paper.

This isn't required for this particular assignment but I'm sure it will help me a lot in the future so thank you so much for this information.

the system needs (1600/12)*48=6400Ah energy to run for 2 days straight. For a battery capable of giving 24Ah I'll need (6400/24)=267 such batteries in parallel.

Yep, that's basically it.

A more practical system would be 12 Rolls-Surrette S2-3560AGM batteries. Six in series to get a 12 volt 3560 amp hour battery, then another series string of six (connected in parallel to the previous six) for a total of 7120 amp hours.

Yes I understand. I just got confused when you said " since your discharge will be a 48 hour discharge (which is greater than 20 hours) the 20 hour number applies." At first I too thought I would get 0.5A.

Yes I understand. I just got confused when you said " since your discharge will be a 48 hour discharge (which is greater than 20 hours) the 20 hour number applies." At first I too thought I would get 0.5A.

Thank you so much to everyone who helped.

Actually, it was billvon who said that (I avoided introducing this idea so early on to avoid muddying the waters). But now that you have advanced to a good level of understanding, you will be able to see through the mud. Let's take a look at what underpins billvon's statement.

Every power source, batteries or no, has an in-built resistance. This resistance explains why you cannot take energy out of the battery at an infinite rate, and also explains why the apparent capacity is a function of the current drain -- some of the battery's energy is lost as heat in this internal resistance. But as the current drain goes down, the power lost in the resistor drops quickly (as the square of the current), so eventually you approach an amp-hour plateau that represents the battery's inherent capacity.

In the example above, discharging a battery over 48 hours necessarily implies a lower current drain than discharging that same battery over 20 hours, so the capacity of the battery should be at least 24Ah, and perhaps somewhat better. In the absence of additional data, one should conservatively assume that the capacity remains at 24Ah, which is what billvon was trying to convey.

Every power source, batteries or no, has an in-built resistance. This resistance explains why you cannot take energy out of the battery at an infinite rate, and also explains why the apparent capacity is a function of the current drain -- some of the battery's energy is lost as heat in this internal resistance. But as the current drain goes down, the power lost in the resistor drops quickly (as the square of the current), so eventually you approach an amp-hour plateau that represents the battery's inherent capacity.

In the example above, discharging a battery over 48 hours necessarily implies a lower current drain than discharging that same battery over 20 hours, so the capacity of the battery should be at least 24Ah, and perhaps somewhat better. In the absence of additional data, one should conservatively assume that the capacity remains at 24Ah, which is what billvon was trying to convey.